Question

Suppose that X is uniformly distributed on the interval [3,8], and that given X=x, Y is uniformly distributed on the interval [0,x]. That is, the conditional PDF of Y given X=x is

fY|X(y|x) =1/x, 0≤y≤x.

The PDF fY(y) of Y is:

fY(y)=⎨(1/5)ln(8/3)), y∈[0,3]
(1/5)ln(8/y), y∈[3,8]
0, otherwise

Calculate P(|X−Y|≤1)
Calculate P(min(X,Y)≤6)

Answer 1

To calculate P(|X−Y|≤1), we first need to find the joint PDF of X and Y. Since X and Y are dependent, we need to use the conditional PDF to calculate the joint PDF:

fX,Y(x,y) = fY|X(y|x) * fX(x)

The marginal PDF of Y can then be found by integrating the joint PDF over all possible values of X:

fY(y) = ∫[3,8] fX,Y(x,y) dx

Given that fY|X(y|x) = 1/x for 0≤y≤x, and X is uniformly distributed on [3,8], we have:

fX(x) = 1/(8-3) = 1/5

Substituting these values into the joint PDF equation, we have:

fX,Y(x,y) = (1/x) * (1/5) = 1/(5x)

To find the marginal PDF of Y, we integrate the joint PDF over all possible values of X:

fY(y) = ∫[3,y] (1/(5x)) dx + ∫[y,8] (1/(5x)) dx
= (1/5) ln(y/3) + (1/5) (ln(8)-ln(y))
= (1/5) (ln(8/3) + ln(y/3))
= (1/5) ln(8y/9)

Now we can calculate P(|X−Y|≤1):

P(|X−Y|≤1) = P(X-Y ≤ 1) + P(Y-X ≤ 1)

For X-Y ≤ 1, we have X ≤ Y+1, so the limits of integration for X will change depending on the value of Y:

P(X-Y ≤ 1) = ∫[0,2] ∫[Y+3,Y+4] (1/(5x)) dx dy + ∫[2,8] ∫[Y+3,Y+4] (1/(5x)) dx dy

Calculating these integrals, we get:

P(X-Y ≤ 1) = (1/5) (∫[0,2] [(ln(Y+4) - ln(Y+3)] dy + ∫[2,6] [(ln(Y+4) - ln(Y+3)] dy + ∫[6,7] [(ln(8) - ln(Y+3)] dy

Simplifying, we get:

P(X-Y ≤ 1) = (1/5) [2ln(3/4) + 4ln(8/9) + (ln(8)-ln(9))]

To calculate P(min(X,Y) ≤ 6), we need to determine the region of the joint PDF where min(X,Y) ≤ 6. This occurs when Y ≤ 6, so we have:

P(min(X,Y) ≤ 6) = P(Y ≤ 6)

Using the marginal PDF of Y that we previously calculated, we integrate from 0 to 6:

P(Y ≤ 6) = ∫[0,6] (1/5) ln(8y/9) dy

Calculating this integral, we get:

P(Y ≤ 6) = (1/5) [6ln(4/3) - 9ln(8/9)]

I hope this helps! Let me know if you have any further questions.